- A$1,7$
- B$5,2$
- ✓$7,1$
- D$3,2$

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$\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}_2 \xrightarrow[\text { (iii) } \mathrm{HBr}{(iv) \mathrm{Mg}, ether, then \mathrm{HCHO} / \mathrm{H}_3 \mathrm{O}^{+}}]{{(i)BH_3}{\text { (ii) } \mathrm{H}_2 \mathrm{O}_2,{ }^{\text {(-) }} \mathrm{OH}}} \mathrm{A}$

$P.$ $C{H_3} - CH = C{H_2}\xrightarrow{{HCl}}$
$Q.$ $C{H_3} - CH = C{H_2}\xrightarrow[{{\text{Peroxide}}}]{{HCl}}$
$\Delta_{ I } G ^{\circ}=-9.478\, kJ\, mol ^{-1}$
If we start the reaction in a closed container at $495\, K$ with $22$ millimoles of $A ,$ the amount of $B$ is the equilibrium mixture is millimoles ..............
(Round off to the Nearest Integer).
$\left[ R =8.314 J mol ^{-1} K ^{-1} ; \ell n 10=2.303\right]$